Question: Solve for $x$ : $3\sqrt{x} + 6 = 9\sqrt{x} + 4$
Explanation: Subtract $3\sqrt{x}$ from both sides: $(3\sqrt{x} + 6) - 3\sqrt{x} = (9\sqrt{x} + 4) - 3\sqrt{x}$ $6 = 6\sqrt{x} + 4$ Subtract $4$ from both sides: $6 - 4 = (6\sqrt{x} + 4) - 4$ $2 = 6\sqrt{x}$ Divide both sides by $6$ $\frac{2}{6} = \frac{6\sqrt{x}}{6}$ Simplify. $\dfrac{1}{3} = \sqrt{x}$ Square both sides. $\dfrac{1}{3} \cdot \dfrac{1}{3} = \sqrt{x} \cdot \sqrt{x}$ $x = \dfrac{1}{9}$